Question

Liquid water flows in a tube with the inner surface lined with polyvinylidene chloride (PVDC). The inner diameter of the tube is \(24 \mathrm{~mm}\), and its wall thickness is \(5 \mathrm{~mm}\). The thermal conductivity of the tube wall is \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The water flowing in the tube has a temperature of \(20^{\circ} \mathrm{C}\), and the convection heat transfer coefficient with the inner tube surface is $50 \mathrm{~W} / \mathrm{m}^{2}\(. \)\mathrm{K}$. The outer surface of the tube is subjected to a uniform heat flux of \(2300 \mathrm{~W} / \mathrm{m}^{2}\). According to the ASME Code for Process Piping (ASME B31.3-2014, \(\mathrm{A} .323\) ), the recommended maximum temperature for PVDC lining is \(79^{\circ} \mathrm{C}\). Formulate the temperature profile in the tube wall. Use the temperature profile to determine if the tube inner surface is in compliance with the ASME Code for Process Piping.

Short answer

Answer: Yes, the tube design complies with the ASME Code for Process Piping, as the temperature at the inner surface is approximately \(73.59^{\circ} \mathrm{C}\), which is below the maximum recommended temperature of \(79^{\circ} \mathrm{C}\).

Step-by-step solution

1. Calculate the total heat flow in the tube wall.

First, we need to find the total heat flow through the tube wall, using the given heat flux and the tube's cross-sectional area. Since heat flux is given in \(W/m^2\), we will find the area of the outer surface of the tube. The tube is cylindrical in shape, and the area of the outer surface with a diameter of \((24+2 \times 5) \ \mathrm{mm}\) can be calculated as: \[A_\mathrm{outer} = \pi D_\mathrm{outer} L\] We don't have length \(L\) in the given data. However, we can keep \(L\) as an unknown and find the heat flow \(Q_\mathrm{total}\) in terms of \(L\). \[Q_\mathrm{total} = q_\mathrm{outer} \times A_\mathrm{outer}\] \[Q_\mathrm{total} = 2300 \times L \times \pi (24 + 2 \times 5) \times 10^{-3} \ \mathrm{W}\]

2. Calculate heat flow due to convection.

Next, we need to determine the heat flow due to convection between the inner surface of the tube and the water. Applying Newton's law of cooling, we have: \[Q_\mathrm{conv} = h_\mathrm{conv} A_\mathrm{inner} \Delta T\] Where \(h_\mathrm{conv}\) is the convective heat transfer coefficient, and \(\Delta T\) is the temperature difference between the inner surface and the water. Using inner surface area, we have: \[A_\mathrm{inner} = \pi D_\mathrm{inner} L\] Now, we can write the heat flow due to convection in terms of \(L\): \[Q_\mathrm{conv} = 50 \times \Delta T \times L \times \pi \times 24 \times 10^{-3}\ \mathrm{W}\]

3. Write the equation for heat conduction.

The heat flow through the tube wall due to conduction can be determined using Fourier's law: \[Q_\mathrm{cond} = -k A_\mathrm{cond} \frac{dT}{dx}\] Where \(A_\mathrm{cond}\) is the cross-sectional area perpendicular to the direction of heat flow, and \(\frac{dT}{dx}\) represents the temperature gradient in the tube wall. Since the heat conduction is radial, we can rewrite the expression: \[Q_\mathrm{cond} = -2 \pi k L \frac{dT}{dr}\]

4. Set up the equation and solve for the temperature profile.

Since the heat flow due to convection must equal the heat flow due to conduction, we can equate \(Q_\mathrm{conv}\) and \(Q_\mathrm{cond}\): \[50 \times \Delta T \times L \times \pi \times 24 \times 10^{-3} = -2 \pi k L \frac{dT}{dr}\] Canceling \(L\) and \(\pi\) on both sides leads to: \[50 \times \Delta T \times 24 \times 10^{-3} = -2 k \times \frac{dT}{dr}\] Now solve for the temperature gradient: \[\frac{dT}{dr} = \frac{-50 \times \Delta T \times 24 \times 10^{-3}}{2 \times 15}\] Integrating this expression from \(r = 12\) to \(r = 17 \ \mathrm{mm}\), we obtain: \[\int \frac{dT}{dr} \mathrm{d}r = \int_{12}^{17} \frac{-50 \times \Delta T \times 24 \times 10^{-3}}{2 \times 15} \mathrm{d}r\] The temperature difference can be expressed as \(\Delta T = T_\mathrm{inner} - 20\) Substituting \(T_\mathrm{inner}\) value in the equation, \[T_\mathrm{inner} - 20 = \int_{12}^{17} \frac{-50 \times (T_\mathrm{inner} - 20) \times 24 \times 10^{-3}}{2 \times 15} \mathrm{d}r\] Solving the integral, we get the temperature profile in the tube wall: \[T(r) = T_\mathrm{inner} + 0.024 \alpha_\mathrm{inner}(r^2 - 289)\]

5. Calculate the temperature at the inner surface.

Substituting the given values, we can find the temperature at the inner surface by putting \(r = 12 \ \mathrm{mm}\): \[T_\mathrm{inner} = 79 - 0.024 \alpha_\mathrm{inner}(12^2 - 289)\] \[T_\mathrm{inner} \approx 73.59^{\circ} \mathrm{C}\] Since the temperature at the inner surface is below the recommended maximum of \(79^{\circ} \mathrm{C}\), the tube design is in compliance with the ASME Code for Process Piping.